Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 17 - Sections 17.1-17.9 - Exercises - Problems by Topic - Page 853: 34d

Answer

$\Delta S_{sys}$ is positive. $\Delta S_{surr}$ is negative. The reaction is spontaneous at high temperatures.

Work Step by Step

The number of moles of gas increases from 1 to 2 and therefore entropy of the system increases. $\Delta S _{sys}$ is positive. Given that $\Delta H^{\circ}_{rxn}=+393.5 \,kJ$, we have $\Delta S_{surr}=\frac{-\Delta H_{sys}}{T}\lt0$ If the reaction is spontaneous, $\Delta G\lt0$ $\implies \Delta H_{sys}-T\Delta S_{sys}\lt0$ Since both $\Delta H_{sys}$ and $\Delta S_{sys}$ are positive, $T$ should be high for $\Delta G$ to be negative.
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